Problem: Is ${293319}$ divisible by $9$ ?
Solution: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {293319}= &&{2}\cdot100000+ \\&&{9}\cdot10000+ \\&&{3}\cdot1000+ \\&&{3}\cdot100+ \\&&{1}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {293319}= &&{2}(99999+1)+ \\&&{9}(9999+1)+ \\&&{3}(999+1)+ \\&&{3}(99+1)+ \\&&{1}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {293319}= &&\gray{2\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{3\cdot999}+ \\&&\gray{3\cdot99}+ \\&&\gray{1\cdot9}+ \\&& {2}+{9}+{3}+{3}+{1}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${293319}$ is divisible by $9$ if ${ 2}+{9}+{3}+{3}+{1}+{9}$ is divisible by $9$ Add the digits of ${293319}$ $ {2}+{9}+{3}+{3}+{1}+{9} = {27} $ If ${27}$ is divisible by $9$ , then ${293319}$ must also be divisible by $9$ ${27}$ is divisible by $9$, therefore ${293319}$ must also be divisible by $9$.